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Is the Square Root of i^4 Equal to 1 or -1?

Date: 02/24/2004 at 08:51:26
From: Mary
Subject: complex numbers

If you take the square root of i to the fourth power, does that equal 
i to the second power, which is equivalent to -1?  Or can you simplify 
under the radical first and say i to the fourth power is 1 and the 
square root is then 1?  Which approach is correct? 

Date: 02/24/2004 at 10:37:01
From: Doctor Jacques
Subject: Re: complex numbers

Hi Mary,

I think the following answer from the Dr. Math archive will tell you 
what you want to know: 

I found it by searching on these keywords:

  square root complex

You can find other similar answers by doing the same search at 

I hope this helps.  Let me know if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum 

Date: 02/24/2004 at 10:42:12
From: Doctor Rob
Subject: Re: complex numbers

Thanks for writing to Ask Dr. Math, Mary!

When you are dealing with roots of complex numbers, you must always
take into consideration the multiple values which may be obtained.  
The square roots of i^4 are 1 and -1.  On the other hand, if you take
the square roots of i, you get (1 + i)/sqrt(2) and -(1 + i)/sqrt(2),
and the fourth power of both of these is -1.  These answers are not
the same, but they do have a certain consistency.

From this you can see that the rules you learned for working with
radicals of nonnegative real numbers, such as

   sqrt(a*b) = sqrt(a)*sqrt(b)   (a > 0, b > 0),
   sqrt(a/b) = sqrt(a)/sqrt(b)   (a > 0, b > 0),

do not necessarily hold for complex numbers or even negative real 
ones.  (Try various combinations of 1 and -1 for a and b to see what I 

For positive real numbers a, the symbol sqrt(a) is defined to be the
positive square root of a.  We call this the "principal value" of the
square root of a.  This is done partly to make the above rule work.  

For negative real numbers and complex numbers, there is no way to pick
one of the two square roots in a consistent way as the principal value
to make the above rules work.  The best you can say is that whichever
choices you make for square roots of such numbers,

   sqrt(a*b) = +/- sqrt(a)*sqrt(b),
   sqrt(a/b) = +/- sqrt(a)/sqrt(b)

with the double sign symbol +/-.

Sorry, but negative real numbers and complex numbers are just not as
simple as positive real numbers when interacting with square root

For nth roots of nonzero numbers, there are always n, of which at 
least n - 2 are complex.  When a is real and positive, there are two
real roots, and you can pick the positive real nth root to be the
principal value.  When a is real and negative, and n is odd, there
is just one real root which is negative, and you can pick that 
negative real nth root to be the principal value.  In all other cases, 
none of the nth roots are real numbers, and there is no good way to 
pick a principal value of the nth root of a in order to make the 
analogue of the first equation above work.

When there is no principal value which can be assigned, the best one
can do for the analogues of the above rules is to write

   (a*b)^(1/n) = w*a^(1/n)*b^(1/n),
   (a/b)^(1/n) = w*a^(1/n)/b^(1/n)m

where w may be any one of the n nth roots of 1 in the complex numbers.

Feel free to reply if I can help further with this question.

- Doctor Rob, The Math Forum 
Associated Topics:
High School Imaginary/Complex Numbers
High School Square & Cube Roots

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